A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod.
1 answer:
Answer:
a = 17.68 m/s²
Explanation:
given,
length of the string, L = 0.8 m
angle made with vertical, θ = 61°
time to complete 1 rev, t = 1.25 s
radial acceleration = ?
first we have to calculate the radius of the circle
R = L sin θ
R = 0.8 x sin 61°
R = 0.7 m
now, calculating at the angular velocity
ω = 5.026 rad/s
now, radial acceleration
a = r ω²
a = 0.7 x 5.026²
a = 17.68 m/s²
hence, the radial acceleration of the ball is equal to 17.68 rad/s²
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Answer:
θ = 62.72°
Explanation:
The projectile describes a parabolic path:
The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).
The equation of uniform rectilinear motion (horizontal ) for the x axis is :
x = x₀+ vx*t Formula (1)
vx = v₀x
Where:
x: horizontal position in meters (m)
x₀: initial horizontal position in meters (m)
t : time (s)
vx: horizontal velocity in m/s
v₀x: Initial speed in x in m/s
The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:
y= y₀+(v₀y)*t - (1/2)*g*t² Formula (2)
vfy= v₀y -gt Formula (3)
Where:
y: vertical position in meters (m)
y₀ : initial vertical position in meters (m)
t : time in seconds (s)
v₀y: initial vertical velocity in m/s
vfy: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
Data
v₀ = 120 m/s , at an angle θ above the horizontal
v₀x= 55 m/s
x-y components of the initial velocity ( v₀)
v₀x = v₀*cosθ Equation (1)
v₀y = v₀*sinθ Equation (2)
Calculating of the angle θ
We replace data in the Equation (1)
55 = 120*cosθ
cosθ = 55 / 120
θ = 62.72°